Make Sum Divisible by P
给一个数组nums和一个p, 求删去最小的子数组后的数字和可以整除p. let s = sum of nums, r = s mod p -> r = (s + p) mod p 因为我们知道presum可以用两个指针表示其中间隔的和. let (i, j) = indexs in presum where 0 <=i <= j <= presum.length. since r = (s + p) mod p, then r = (pre[j] – pre[i] + p) mod p 因为我们知道2sum中, […]